Example Calculation - 4

This worked example uses uncertainty to evaluate an experimental result.Question: in an experiment to measure the enthalpy change of neutralisation, the temperature of the mixture rose from 19.5 °C to 26.0 °C. The thermometer is read to ± 0.5 °C. The student's value was −45 kJ mol^-1; the data book value is −57 kJ mol^-1. Can the difference be explained by the uncertainty in the temperature measurement?Step 1: percentage uncertainty in the temperature rise.temperature rise = 26.0 − 19.5 = 6.5 °C, with uncertainty 0.5 + 0.5 = ± 1.0 °Cpercentage uncertainty = (1.0 ÷ 6.5) × 100 = 15.4%Step 2: percentage difference from the accepted value.percentage difference = ((57 − 45) ÷ 57) × 100 = 21%Step 3: compare. The difference (21%) is larger than the measurement uncertainty (15.4%, the largest source of uncertainty here). So the uncertainty alone cannot explain the result, and there must also be a systematic error.Evaluation: the result is less exothermic than expected, which fits heat being lost to the surroundings, a systematic error that makes every temperature rise too small. Improvements: insulate the cup and fit a lid, and use a thermometer read to ± 0.1 °C to cut the percentage uncertainty.The general rule: if the percentage difference is within the total percentage uncertainty, the method is acceptable and the difference is due to the apparatus; if it is greater, look for systematic errors in the method.